阈值博弈：最大化期望收益

选择阈值 $c\\ge 0$，若 $U>c$ 得到 $c$，否则 0。期望收益\n\n$$\nG(c)=c\\,[1-F(c)].\n$$\n\n一阶条件\n\n$$\nG'(c)=1-F(c)-c f(c)=0\\Rightarrow \\boxed{1-F(c)=c f(c)}.\n$$\n\n若 $U\\sim\\mathrm{Exp}(\\lambda)$，则 $1-F(c)=e^{-\\lambda c}$、$f(c)=\\lambda e^{-\\lambda c}$，得到 $\\lambda c=1$，所以\n\n$$\n\\boxed{c^*=\\frac{1}{\\lambda}},\\\qquad \\boxed{G(c^*)=\\frac{1}{e\\lambda}}.\n$$

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英文解析

a. As stated in the Hints, in order to win a large amount, one would like to choose a rather large value of c. On the other hand, the larger c is chosen, the smaller the probability of winning anything at all. Let us call the chosen value of c the "strategy"of the player and denote its expected gain as G(c). Thus, with the strategy c, the gain will be zero with probability $\mathrm{P}(\mathrm{U} \mathrm{S} \mathrm{c}) = \mathrm{F}(\mathrm{c})$ , and it will be equal to c with probability $\mathrm{P}(\mathrm{U} > \mathrm{c}) = 1 - \mathrm{F}(\mathrm{c})$ . Putting these two cases together, we get

$$G(c) = 0\cdot F(c) + c\cdot [1 - F(c)] = c\cdot [1 - F(c)]$$

Figure 6.1: The relation between the expected gain, G (c), and the number, c, that was chosen. The example assumes that U has an exponential distribution f $(\mathbf{t}) = \lambda$ exp $(- \lambda t)$ with mean $1 / \mathrm{A} = 50.$ The optimal choice of c is then just 50, in which case the expected gain equals 18.39.

To get an idea about G, it is useful to first study its boundary values. From the "tail formula" (see page 2) for the expected square of U,

$$\begin{array}{l}{\operatorname {E}[\mathbf{U}^{2}] = \mu^{2} + \sigma^{2}}\\ {= 2\int_{0}^{\infty}u[1 - F(u)]du}\\ {= 2\int_{0}^{\infty}G(u)du< \infty} \end{array}$$

because the first two moments of U were assumed to be finite. It must then be the case that $\mathbf{G}(\mathbf{c})$ tends to zero as $\mathbf{c} \to 8$ ; otherwise the integral would diverge. Also, G is nonnegative, and from $\mathbf{G}(0) = \mathbf{G}(\mathbf{o}) = 0$ we conclude that G must have at least one maximum over the positive reals.

Differentiating G with respect to c and setting this derivative equal to zero gives

$$\begin{array}{c}\frac{dG(c)}{dc} = [1 - F(c)] - cf(c) = 0\\ \mathrm{i.e.,}\quad \frac{f(c)}{1 - F(c)} = \frac{1}{c} \end{array}$$

b. The left-hand side of the last equation is the hazard function h of U; cf. page 2. The equation states that the optimal strategy c is the abscissa at which the hazard function h(c) crosses the function $1 / c$ .

c. For an exponential rv, the hazard function is constant, $\mathbf{h}(\mathbf{c}) = \mathbf{A}$ . There- fore, the optimal strategy is $\mathbf{c} = 1 / \mathbf{A}$ , i.e., to choose the expected value of U, see Figure 3.5. With this strategy, the expected gain is

$$G(\frac{1}{\lambda}) = \frac{1}{\lambda}\cdot e^{-1}$$

i.e., about $36.8\%$ of the expected value of U. This value appears remarkably small, but due to the large spread (relative to its mean) of the exponential distribution, there is no way to further improve it within the given setting.